# Calculus II

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Here is the picture I forgot earlier.

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The formula you need is

`S_x=2pi int_a^by (ds)/dx dx`

So in your case you have

`S_x=2piint_0^pi k(sint-tcost)(ds)/dx dx`

Since your curve is given parametrically you have `ds=sqrt((dx/dt)^2+(dy/dt)^2)dt`

`S_x=2kpiint_0^pi(sint-tcost)sqrt([k(-sint+sint+tcost)]^2+[k(cost-cost+tcost)]^2)dt=`

`2kpiint_0^pi(sint-tcost)sqrt(k^2t^2(cos^2t+sin^2t))dt=`

since `cos^2t+sin^2t=1` we have

`2kpiint_0^pi(sint-tcost)ktdt=`

`2k^2piint_0^pi(tsint-t^2cost)dt`

Now we have two integrals for which we will use integration by parts

First

`I_1=inttsintdt=|(u=t,dv=sintdt),(du=dt,v=-cost)|=-tcost+intcostdt=`

`-tcost+sint+C`

Similarly (we need to apply integration by parts two times) we obtain solution for the second integral.

`I_2=int t^2costdt=2tcost+(t^2-2)sint+C`

Now we combine those two integrals (add them)

`I_1+I_2=-3tcost-(t^2-3)sint`

and return back to calculation the surface area.

`S_x=2k^2pi(-3tcost-(t^2-3)sint)|_0^pi=`

`2k^2picdot3pi=6k^2pi^2`